Organic Chemistry Dr V.O. Nyamoricheminnerweb.ukzn.ac.za/Files/Chem120 Organic 2010... · 7/11/2010 1 Organic Chemistry Dr V.O. Nyamori Textbook by i)Bruice “Organic Chemistry”

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Organic Chemistry Dr V.O. Nyamori

Textbook by i) Bruice “Organic Chemistry” 5th Edition

ii) Hart et al. “Organic Chemistry: A Short

Course” 12th Edition

• The study of carbon‐containing compounds and their properties.

• The vast majority of organic compounds contain chains or rings

of carbon atoms.

• They form the basis of, or are important constituents of many

products (plastics drugs petrochemicals food explosives

Course 12 Edition

products (plastics, drugs, petrochemicals, food, explosives,

paints, to name but a few) and, with very few exceptions, they

form the basis of all earthly life processes.

1

Soap/detergent ‐surfactant C17H35COO‐

Sugar ‐glucose C6H12O6

Medicine ‐ ascorbic acid HC6H7O6

e.g.

Carbon: group 14, atomic no. 6Recall: Electronic configuration for carbon?

Periodic Table

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Structure of Carbon Compounds

Three hybridization states

C C C C C C

1 54 Å 1 20 Å1 33 Å

each satisfies the octet rule for each carbon!!

1.54 Å 1.20 Å1.33 Å

Hart et al. “Organic Chemistry: A Short Course”, 12th Edition,

Chapter 1.14 ‐ 1.18

Geometries of carbon compounds

sp3 Tetrahedral 108.5°

sp2 Trigonal planar 120°

sp Linear 180°

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Hydrogens are in a tetrahedral

arrangement around the sp3

hybridized carbon atom

Methane: CH4

hybridized carbon atom.

Hydrogens bond to the carbon sp3

orbitals with 1s orbitals.

sp3 Hybridizationcarbon

Energy

sp3

2p

2s

1s

Hybridization

nergy

2p

2s H b idi ti

sp2 Hybridizationcarbon

En

sp22s

1s

Hybridization

2p

Ethene: C2H4

sp Hybridizationcarbon

sp

p

2s

1s

Hybridization

Energy

Ethyne: C2H2

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HYDROCARBONS

• compounds composed of only carbon and hydrogen

• chain of carbon atoms bonded to enough hydrogen atoms

to satisfy the octet rule for each carbonto satisfy the octet rule for each carbon

• chain is bent because of the 109.5° C–C–C tetrahedral angle

CCH

H

HH

CH

CHHC

CH2

CC

H3C

e.g.

C

CCC

CC

H

H

H

H

HH

H CH3

Line notation

HYDROCARBONS

• Hydrocarbons with all single carbon‐carbon bonds(no double or triple bonds)

Alkanes

• They contain the maximum number of hydrogen atoms

(no double or triple bonds)

• Alkanes are SATURATED

Alkenes, alkynes and aromatic compounds

8

• UNSATURATED hydrocarbons

‐ they ARE NOT ALKANES

• contain carbon‐carbon multiple bonds

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Past exam Question

C‐1: _____ C‐3: _____

C‐2: _____ C‐4: _____

Example

HBr

H

1. Indicate the hybridization for carbons 1 – 10 and theirrespective geometry. Include bond angles in your answer.

CC

C

CC

CC

C

O

CN

F H

O

ClH

H

H

H

12

3

46

78

910

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Hydrocarbons

• Four basic types:‐ Alkanes

CnH2n+2

‐ Alkenes

Alkynes

C2H6 Ethane

CnH2n

C2H4 Ethene 120°

‐ Alkynes

‐ Aromatic hydrocarbons

C2H2 Ethyne

CnHn

CnH2n

HH

H

H H

H

C6H6 Benzeneor

Organic Nomenclature

• Three parts to a compound name:

1 2 3

1. Prefix

2. Base/parent

3. Suffix

Chapter 2: “Organic Chemistry” 5th Edition , Bruice P. Y.

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Organic Nomenclature ‐ IUPAC Rules

Suffix: Tells what type of compound it is.

Prefix : Tells what substituent(s) are attached, if any.

prefix base suffix

Base/parent: Tells how many carbons are in the longest continuous chain.

prefix base suffix

What substituent? How many

carbons?

What family?

Alkanes

• Only van der Waals force: London force.

• Boiling point increases with length of chain.

• Nomenclature suffix “‐ane”

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To Name a Compound…1. Determine what type of

compound it is.

2. Find the longest chain in the molecule.

3. Number the chain from the end nearest the first substituent encountered.

4. List the substituents as a prefix in alphabetical orderprefix in alphabetical order along with the number(s) of the carbon(s) to which they are attached.

CH2CH

CH2H2C

CH2

H3C

H3C

CH3Name??

ExampleCH3

CH2CHCH3

If there is more than one type of substituent in the molecule,

CH CH

CHCH3

CH2 CH3

CH3

CH3

list them alphabetically i.e. name of substituent, not prefix for

frequency e.g. di, tri, tetra, etc...are not considered.

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Cycloalkanes

• Carbon can also form ringed structures.

• Five‐ and six‐membered rings are most stable.

– Can take on conformation in which angles are

very close to tetrahedral angle.

– Smaller rings are quite strained.

l f l

cyclohexane cyclopentane cyclopropane

General formulaCnH2n

CYCLIC ALKANES

How do we name….CH3

CH2CH3

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Alkenes

VSEPR Theory

• Contain at least one carbon–carbon double bond

120°

• Unsaturated

– Have fewer than maximum number of hydrogens

– The C atoms on double bond are sp2 hybridized

Structure of Alkenes

• Unlike alkanes, alkenes cannot rotate

freely about the double bondfreely about the double bond.

– Side‐to‐side overlap makes this

impossible without breaking ‐bond.

C C

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Structure of Alkenes

This creates geometric isomers

diff i h i ldifference in the spatial arrangement of groups about the double bond

Cis‐ isomer “Z”‐isomer

Z‐2‐Pentene

Trans‐ isomer “E”‐isomer

E‐2‐Pentene

NAMING ALKENES

1. Find the longest unbranched carbon chain containing the

d bl b ddouble bond.

2. Number the carbon atoms in the main chain.

• Name chain according to number of carbon atoms.

add ‐ene as a suffix

• Start from the end of the chain that is closest to the

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Start from the end of the chain that is closest to the

double bond.

location of the double bond is numbered with the

lowest‐numbered carbon in the double bond.

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Example

H

Name this alkene

C

H

CHCH3 C

H

CH

CH2

CH2

CH

CH3

CH3

H2C

23

Name this alkene

Example

C CHCH3 C

H

CH CH2 CH3

CH3

1. The longest unbranched chain containing the double

bond which is the functional group (suffix ‐ ene)

Name this alkeneH

CH2

CH

H2C

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bond which is the functional group (suffix ‐ ene)

2. The chain numbering starts closest to the double bond.

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3. There are two substituent groups on this alkene:

C CHCH3 C

H

CH CH2 CH3

CH3

4. Compose the name…..

• Add the substituent groups alphabetically to the alkene name

H

CH2

CH

H2C

name of the alkene is?

• Specify the position of each group on the main chain

Properties of AlkenesExample: C4H8

2‐Methyl‐1‐propenebp. ‐7 ⁰C

1‐Butenebp ‐6 ⁰C

Cis‐2‐Butenebp +4 ⁰C

Trans‐2‐Butenebp +1 ⁰C

Structure also affects physical properties of alkenes

Can we have more than one double bond?

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Alkenes exhibit cis‐trans isomerism.

GEOMETRICAL ISOMERS

Trans‐ isomer “E‐” Cis‐isomer “Z‐”

• Identical substituents on

opposite sides of the

double bond

C

H

CCH3

CCH3

CCH3

• Identical substituents on

same side of the double

bond

CH3 HC

HC

Htrans‐“E‐”

cis‐“Z‐”

Stick diagram

Geometric isomers of Alkenes• Cis‐alkenes have similar higher priority elements or group (or

Z‐isomer i.e. higher priority elements but not necessarily

similar) in the chain on the same side of the molecule.

• Trans‐alkenes have similar higher elements or group (or E‐Trans alkenes have similar higher elements or group (or E

isomer i.e. higher priority elements but not necessarily similar)

in the chain on opposite sides of the molecule.

• Priorities are assigned by the atomic numbers of the atoms

bonded to the carbon in the double bond.

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Examples1. Name the following alkenes and determine whether there

are geometric isomers i.e. either Trans‐ (E‐) or Cis‐ (Z‐)isomers.

H

a) Br

Hb)

F

c)

Certain groups of atoms give a molecule a....FUNCTION

Acidic, basic, alcohol, etc…

FUNCTIONAL GROUPS

Each functional group is specified by a suffix or prefix

The GROUPS are called functional groups.

depicted on the nomenclature of the organic molecule

Functional groups are given an order of priority to

30

decide on which is the suffix.

Please refer to your textbooks on priority preference by:

i) Bruice “Organic Chemistry” 5th Edition

ii) Hart et al. “Organic Chemistry: A Short Course” 12th Edition

HOMEWORK!

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Group / Family

FormulaStructural Formula

Prefix Suffix Example

Alkane RH alkyl- -ane

Ethane

Alkene R2C=CR2 alkenyl- -ene

Alkyne RC≡CR' alkynyl- -yne

Ethene

Benzene derivative

RC6H5

RPhphenyl- -benzene

2-phenylpropaneisopropylbenzene

Ethyne

Group / Family



Haloalkane RX halo-alkyl

halide

Fluoroalkane RF fluoroalkyl

Chloroethane

Ethyl chloride

FluoromethaneFluoroalkane RF fluoro-

fluoride

Chloroalkane RCl chloro-alkyl

chloride

Bromoalkane RBr bromo-alkyl

bromide

Methyl fluoride

Chloromethane

Methyl chloride

Bromomethane

M th l b idbromide

Iodoalkane RI iodo-alkyl

iodide

Methyl bromide

Iodomethane

Methyl iodide

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Group / Family

Group FormulaStructural Formula


Primary amine

RNH2 amino- -amine

Methylamine

Amines

Secondary amine

R2NH amino- -amine

Tertiary amine

R3N amino- -amine

Dimethylamine

Trimethylamine

Quaternaryammonium

ionR4N+X-

ammonio--ammonium

H3C

N

H CH3

CH3

Cl

Trimethyl-ammonium

chloride

Group / Family



Alcohol ROH hydroxy- -ol

M th l

Ketone RCOR' keto-, oxo- -one

Methanol

ButanoneMethyl-

ethyl ketone

Aldehyde RCHO aldo- -al

EthanalAcetaldehyde

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Group / Family



Carboxylic acid

RCOOH carboxy- -oic acidEthanoic acid

Acetic acid

Acyl halide RCOX haloformyl- -oyl halide

Eth ROR' lkalkyl alkyl

Ethanoyl chlorideAcetyl chloride

Ether ROR' alkoxy-y y

ether

Ester RCOOR'alkyl

alkanoate

EthoxyethaneDiethyl ether

Ethyl butanoateEthyl butyrate

Primary (1°) alcohols and amines

General structure

Alcohol H

Example

HAlcoholR1 C

H

OH

1°

C

H

H

OHCH3CH2

HH

Amine

1°

N

H

H

CH3CH2R1 N

H

H

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Secondary (2°) alcohols and amines

General structure

Alcohol H

Example

HAlcoholR1 C

R2

OH

2° Alcohol

C

H

CH3

OHCH3CH2

HHAmine

2° Amine

N

H

CH3

CH3CH2R1 N

H

R2

Tertiary (3°) alcohols and amines

General structure

Alcohol1

R3

Example

CH3

R1 C

R2

OH

3° alcohols

C

CH3

OHCH3CH2

CH3R3

Amine

3° Amine

N

CH3

CH3

CH3CH2R1 N

R3

R2

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Quaternary amines

General structure

R2 CH

Example

AmineR4 N

R2

R1

R3+ N

CH3

CH3

CH3CH3CH2

+

Exercise

1. Draw the structures of the following alcohols and amines

and classify them as either 1°, 2°, 3° or quaternary

a) Pentan‐1‐ol

b) Dimethylamine

c) 3‐Ethyl‐hexan‐3‐ol

d) Diethyl‐methyl‐amined) Diethyl methyl amine

e) Butan‐2‐ol

f) Triethylmethly ammonium ion

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a) Pentan‐1‐ol

Solutions

b) Dimethylamine c) 3‐Ethyl‐hexan‐3‐ol

d) Diethyl‐methyl‐amine e) Butan‐2‐ol f) Triethylmethly‐ammonium ion

Naming Hydrocarbons with Functional Groups

Name the other substituent groups, using the

prefixes for alkyl groups and the prefixes for any

other functional groups

Specify the position of each group on the main chain.

Add the substituent groups alphabetically to the name

f th lk ( lk lk ) l ith th

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of the alkane (or alkene or alkyne) along with the

frequency of each group.

Example…..

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Example: Name this organic molecule:

CH3 CHCH

OH

CH3

NO2

3 3

43


CH3 CHCH

OH

CH3

NO2

1. This molecule contains a ……………. and …………. group

3 3

Only the group has priority.

44

Only the …………… group has priority.

So this is an ................

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CH CHCH

OH

CH

NO2

longest carbon chain containing the hydroxyl

CH3 CHCH CH3

45

g g y y

group has ………… carbons.

Therefore its a “….……….. ”

An alcohol suffix is ‐OL

CH3 CHCH

OH

CH3

NO2

3. Number the carbons, starting NEAREST the

functional group.

The hydroxyl group is on position …

…so this is a ………… or .…………

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4. This molecule has one substituent

A nitro group on position …

………………………… …………………….or

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Example:

Name this organic

molecule ?? CH CH CH CH C

CH2

CH2C

CH3

CH3

O

H2N

molecule ?? CH3 CH CH CH2

CH3

C

CH2

CH3

OH

1. Identify functional groups

47

2. Determine longest carbon chain that contains the highest priority functional group.


CH

CH2C

CH3

CH3H2N

CH3 CH CH CH2

CH3

C

CH2

CH2

CH3

OH

O

The longest carbon chain has 8 carbons

48

The longest carbon chain has 8 carbons.

BUT………………...

THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group

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Example: Name this organic molecule

CH

CH2C

CH3

CH3H2N

CH3 CH CH CH2

CH3

C

CH2

CH2

CH3

OH

O

The longest carbon CONTAINING the ‐CO2H

49

g 2

chain has …. carbons.

so this molecule is based on a ..………… ..

CH3 CH CH CH2 C

CH2

CH2C

CH3

CH3

O

H2N

Highest priority functional group is a carboxyl group

3 2

CH3 CH2

CH3

OH

Suffix OIC ACID

50

Suffix ‐OIC ACIDHence… Heptanoic acid

Now number chain…….

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CH3 CH CH CH2 C

CH2

CH2C

CH3

CH3

O

H2N

Heptanoic acid

3. Number the carbons starting with the functional group

3 2

CH3 CH2

CH3

OH

The carboxyl group is on position 1The carboxyl group is on position 1,

do not include in the name because

the carboxyl group is always a terminal group.

CH3 CH CH CH2 C

CH2

CH2C

CH3

CH3

O123

4

5 6 7H2N

Heptanoic acid

h

3 2

CH3 CH2

CH3

OH

4. This molecule has four substituents

52

Compose the name

ALPHABETICAL LIST

INTERPETING AN IUPAC NAME…...

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WHAT IS THE STRUCTURAL FORMULA OF

BUTANONE ?

CH3 CH2C

O

CH3

Carbonyl CANNOT Be at the END!! WHY?????

53

C CH2CH2

O

CH3H

BUTANAL

Examples

Give the correct IUPAC name for the following compounds.

Bra) b)) )

H

OH

c) d)

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O

H CH3

Examples

Give the correct IUPAC name for the following compounds.

f) g)H

O

CH3

Cl

)

h) i) O

O

OHh) i) O

O

Examples

Draw structural formulae for the following compounds:

a) 2,3,5‐trimethylhexane

b) (Z)‐3‐chloro‐hept‐2‐ene

c) 3‐ethylnonanol

d) 2,3‐dimethylpentanoic acid

e) methylhexanoatee) methylhexanoate

f) 3‐iodohexanal

g) pentan‐2‐one

h) 3‐aminopentane T1

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Solutions

2,3,5‐trimethylhexane (Z)‐3‐chloro‐hept‐2‐enea) b)

3‐ethylnonanol 2,3‐dimethylpentanoic acidc) d)

Solutions

methylhexanoate 3‐iodohexanale) f)

O OI

pentan‐2‐oneg) h)

O

OI

H

3‐aminopentane)

OH2N

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ISOMERS

(a) Structural isomers

Molecules with the same chemical formula but different

Two types: (a) Structural isomers (b) Stereoisomers

CONSTITUTIONAL ISOMERS

Constitutional isomers have different properties:

e.g. butane (C4H10) has 2 structural isomers

Molecules with the same chemical formula but different

bonds between the atoms

Now called...

These are…..?

CH3 CH2 CH2 CH3 CH3 CH CH3

CH3

n‐Butane: C4H10 2‐methylpropane: C4H10

bp = ‐12 °C mp = ‐159 °Cbp = 0 °C mp = ‐138 °C

Solution:

ExampleHow many constitutional isomers are formed from C5H12? Draw their structures.

60

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CONSTITUTIONAL (STRUCTURAL) ISOMERS

The general formula for ALKANES is…... 22 nnHCn = 1, 2 and 3 1 ISOMER

The number of ISOMERS increases with n…..

,

n = 4 2 ISOMERS

n = 5 3 ISOMERS

n = 6 5 ISOMERS

n = 7 9 ISOMERSn = 40

n = 8 18 ISOMERS

n = 9 35 ISOMERS

n = 10 75 ISOMERS

n = 20 366,319 ISOMERS

n = 40 62,491,178,805,831 ISOMERS

HOMEWORK: DRAW THE ISOMERS OF C40H82 !!!

Constitutional isomers for multibonds

C6H10

Example: Alkene

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Constitutional isomers for multibonds

Example: Alkyne

C6H6

H3C C C C C CH3

C C C C CH2

CH3

H

2,4‐Hexadiyne

Hexa‐2,4‐diyne

63

C C C C CH2H

1,3‐Hexadiyne

Hexa‐1,3‐diyne

OPTICAL ISOMERISM

Optical isomerism arises when molecules have a

structure such that the mirror image is notstructure such that the mirror image is not

superimposable on the original molecule.

occurs whenever there are four different groups

bound to the same tetrahedral carbon atom.

64

Some terminology…….

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OPTICAL ACTIVITY IN ORGANIC COMPOUNDS

Stereogenic centre has four different groups attached

to a tetrahedral carbon atom

ZX

W

Y

C

Stereogenic centre

Chiral carbon atom

*

65

The carbon involved is called a chiral carbon or

stereogenic carbon and the molecule is known as a chiral

molecule.Example: 2‐butanol

Y

2‐Butanol

Dash shows bond going Bonds aligned to

Perspective formula

C

CH CH

OH

H3C

Dash shows bond going backwards from the viewer

*

gthe asymmetric

center in the plane

CH2CH3H

3

Solid wedge represents a bond extending out towards

the viewerChiral centre

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CH

2‐Butanol

Fischer projection

Bonds aligned to Bond going

CH2CH3HO

CH3

*

gthe asymmetric

center in the plane

Bond going backwards from the

viewer

Bond extending d h

H

67

Chiral centre

out towards the viewer

2‐butanol has two optical isomers.

A pair of isomers called enantiomers ‐

ENANTIOMERISM in ORGANIC CHEMISTRY

non‐superimposable mirror images of each other.

View in 3‐DView in 3‐D“mirror”

OH

C

OH

68

C

CH2CH3H

H3CC

HCH3CH3CH2

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Identify the chiral carbon (stereogenic centre) and draw the

structural formula for each of the following molecules:

(a) 1‐chloroethanol (b) 2,3,5‐Trimethyl‐hexane

( ) h l l h (d) di hl l

Question

Solution

(c) Methylcyclohexane (d) 1,3‐dimethlycyclopentane

C CH3HO

H

(a) 1‐chloroethanol

*

*

(b) 2,3,5‐Trimethyl‐hexane

3

Cl

* *No chiral carbon

(d) 1,3‐dimethlycyclopentane(c) Methylcyclohexane

NAMING OPTICAL ISOMERS

Stereogenic centre creates twomolecular optical isomers

Cl ClTwo

configurations

How do we name these isomers??

CH

OHH3C

CH

HO CH3

configurations

Enantiomers

70

Solution: Use R‐S nomenclature system for designating

the configuration

We assign priorities as in the E, Z system……...

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1. Assign relative priorities to each of the four groups on

the stereogenic carbon to describe the configuration.

CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE

The priorities are given by rules:

• Higher atomic numbers are given higher priorities.

• If necessary, the second atom in each substituent is

71

y,

used to determine the priorities.

2. Draw the molecule with the lowest‐priority group

pointing directly into the page….

CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE

and the other three groups pointing out of the page in

an arrangement like a steering wheel.

Example: 2‐ Butanol…..CHCCH

OH

CH

72

CH2CCH3 CH3

H

Draw molecule as a wedge and dashed line diagram

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O O

2‐ butanol

CH2CCH3

OH

CH3

H

p

Hart et al. “Organic

Chemistry: A Short Course” 12th Edition. Chapter 5

H3CCH2CH3

C

OH

HH3CH2C

C

OH

CH3

H

H

73

DO NOT FORGET THE OTHER ISOMER……….

THESE ARE THE TWO…. ENANTIOMERS...or ....OPTICAL ISOMERS

1. Assign priorities:

O > CMe = CEt > H

C,H,H > H,H,H

2‐ butanol

C

CH2CH3

OH

H, , , ,

CEt > CMe

CH3

3

74

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CHCH2CH3

C

OH

H

1

24

O > CEt > CMe > H

1. Assign priorities to each group:

2‐ Butanol

CH33

CH CHC

OH1

2

2. Redraw the molecule with the lowest priority group facing in.

Et Me

75

H3C CH2CH33

Now what?????

H C CH CH

OH

1

2C

2‐ Butanol

3. Look at the direction in which the priorities decrease.

If they decrease in a clockwise direction, the

H3C CH2CH33

(R)‐2‐Butanol

76

stereogenic centre is called “R” or rectus

which is Latin for “right.”Or….

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CH3CH3CH2

OH

1

2 3C

2‐ Butanol

3CH3CH2

If the priorities decrease in a counter‐clockwise

di ti th t i t i ll d “S”

(S)‐2‐Butanol

direction the stereogenic centre is called “S” or

sinister, which is Latin for “left.”

Example:

NH2 NH2

What are the configurations of the following chiral molecules?

CH3

CH2CH3

CH

CH3CCH2CH3

H

78

H3C CH2CH3

C

NH2CH3

CH2CH3

CH2N

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Example: Give the configuration of the stereogenic

centre in each of the following molecules:

CH2OHCl

CH

ClC OH

O

CHCH2OHHO C

O(A) (B)

79

THEY LOOK DIFFERENT BUT ARE THEY???

2‐chloro‐3‐hydroxy‐propanoic acid

A)

CH

CH2OH

ClC OH

O

1. Assign priorities to each group:

O

ClC

CH2OH

C OH

2. Redraw the molecule with the

lowest priority group facing in.

3. Look which way the priorities decrease

8080

C OH

O

The priorities decrease anti‐clockwise, so this centre is “.…”

decrease.

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1. Assign priorities to each group

CHCH2OH

ClB)

2. Redraw the molecule with CH2OHHO C

Othe lowest priority group

facing in.

C

Cl 3. Direction of the priorities

decrease?C

CH2OHHO C

O

The priorities decrease anti‐clockwise, so this

centre is “…”

Example: Give the configuration of the stereogenic

centre in each of the following molecules:

ClCH2OH B)A)


CHCH2OHHO C

O

CH

ClC OH

O

82

THEY LOOK DIFFERENT BUT ARE THEY???


WHAT IF?????

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If we flip the COOH and H ???????

CH2OHA) CH2OHA’)

CH

ClC OH

O

C H

Cl

C

O

HO

CH2OH CH2OH

ClC

C OH

O

CC

O

HO Cl

What is the structural formula of (R)‐2‐chloro‐2‐butanol?

DRAW MOLECULAR STRUCTURE

EXAM QUESTION

Solution:

84

ASSIGN PRIORITIES

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(R)‐2‐chloro‐2‐butanolAssign priorities

Draw steering wheel with lowest priority group pointing in…….

MAKE SURE PRIORITY GOES

85

DRAW MOLECULE…..

MAKE SURE PRIORITY GOES CLOCKWISE FOR R

C

DRAW MOLECULE TO SEE ALL GROUPS…..C

Remember lowest priority group h “d h d” b d

Make sure priority goes clockwise for “R”

C

has a “dashed” bond

C

(R)‐2‐chloro‐2‐butanol

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[2R,3R] [2S,3S] [2R,3S] [2S,3R]

a b c dCH CH

CH3 CH3

CHCH3 CH

Br Cl

CH3* *

R

R

C

C

CH3

CH3

H

Cl

Br

H

S

S

C

C

CH3

CH3

Br

H

H

Cl

R

S

C

C

CH3

H

H

Br

ClR

SC

C

CH3

Br

Cl

H

H

enantiomeric pairs enantiomeric pairs

• Enantiomeric pair differ only in optical activity

RCH3

SCH3

RCH3

SCH3a b c d

Diastereomers b & ca & c

b & da & d

R

S

C

C

CH3

H

H

Br

ClR

SC

C

CH3

Br

Cl

H

H

R

R

C

C

CH3

H

Cl

Br

H

S

S

C

C

CH3

Br

H

H

Cl

Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp., , solubility etc…

molecule with n stereogenic centres may exist inmaximum of 2n stereisomeric forms, with maximum of2n/2 enantiomeric pairs

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Meso compoundCHHOC CH COH

OH OH

OO* *

RCO2H

SCO2H

RCO2H

SCO2H

R

R

C

C

CO2H

OH

H

H

HO

S

S

C

C

CO2H

H

OH

HO

H

R

S

C

C

CO2H

OH

OH

H

HR

SC

C

CO2H

H

H

HO

HO

170 °C 170 °C 140 °C

+12° ‐12° 0°

meso compound ‐ an achiral (optically inactive)

diastereomer of compound with stereogenic centres

arises because 4 different groups making each of C‐2 & C‐3

stereogenic are same 4 different groups…(!)

Meso compound

RC

CO2H

OHHS

C

CO2H

HHOR

C

CO2H

OHHS

C

CO2H

HHO

RC

CO2H

HHOS

C

CO2H

OHHS

C

CO2H

OHHR

C

CO2H

HHO-------------------- -------------------- -------------------- --------------------

Enantiomers, Chiral Identical, achiral, Meso form

possess plane of symmetry bisecting central C‐C bond

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Shows arrangements in space

Newman projections

e.g. Ethane C2H6

H H HH

60° Staggered H H

HH

HH

H

H

“dash‐wedge”

H

HH

HH

H

“sawhorse”

0°

Newman

Staggeredconformation

Eclipsed

H

H

H

H

H

H H

HHH H

“dash‐wedge”

H

HHH

H H“sawhorse”

0

Newman

Eclipsed conformation

HH

HHH

H

a

e

e aa

e "flip"16

5

3

a

e

e

a e

aa

e

e

16

5

43

2

Cyclohexane conformations

ae

a

e

a

e

42

aa

e

H

HH

Chair conformations

H H

HH

H

HH

H

H

HBoat conformation

Cyclohexane

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Examples1,2‐Dimethylcyclopentane

CH3 CH3 H

H

H CH3 H H

H

H

H H

H

H H

3

CH3H

H H

H

H

H H

Different bond pattern

Structural (constitutional)

Summary of isomerism

Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54

pattern

isomers

( )isomer

Stereoisomer

Same bond pattern

Interconvertible by single bond rotation

Not interconvertibleby bond rotation

Conformers (rotamers)

Configurational isomers

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Exercise

1. Draw the structures of

a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐Heptadiene

b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐Bromobutan‐2‐ol

2. Using the Newman projection draw the structure a

staggered conformation of butane.

3. Using the Fischer projection draw a structure of

(S)‐2‐methylpentanoic acid.

4. Draw the structure for the cis and trans isomers of

1‐bromo‐2‐chlorocyclopropane

Solutions

1. a) (Z)‐3‐methyl‐2‐pentene b) (S)‐2‐bromopropanol

c) (E,Z)‐2,4‐Heptadiene d) (2S,3R)‐3‐Bromobutan‐2‐ol

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Staggered conformation of butane

Solutions

2.

Newman projectionp j

3. (S)‐2‐methylpentanoic acid

Fischer projection

(4) 1‐bromo‐2‐chlorocyclopropane

Documents

Organic Chemistry Dr V.O. Nyamoricheminnerweb.ukzn.ac.za/Files/Chem120 Organic 2010... · 7/11/2010 1 Organic Chemistry Dr V.O. Nyamori Textbook by i)Bruice “Organic Chemistry”